91Ó°ÊÓ

Differentiation is a fundamental concept in calculus. It involves finding the rate at which a function is changing at any given point, which is integral when you want to find the slope of a curve. In our exercise, we began by differentiating the function representing the curve. The original function given was \( y = 4 - x^2 \).

To find the derivative, we need to apply a set of rules in calculus. The main rule used here in the solution is the power rule. Differentiation provides us with a new function that describes how \( y \) changes with respect to \( x \). In simpler words, it helps us find the 'slope function' for the curve. Once you've taken the derivative, it becomes easier to calculate the slope of the tangent at any specific point. In this problem, this "slope function" is \( \frac{dy}{dx} = -2x \). This function tells us that at any point \( x \) on the curve, the slope of the tangent line is simply \( -2x \).

The power rule is a quick and efficient method for finding the derivative of a function of the form \( x^n \). In our exercise, the function \( y = 4 - x^2 \) includes the term \( x^2 \), and the power rule helps to differentiate this term effortlessly.

The power rule states that if \( y = x^n \), then the derivative of \( y \) with respect to \( x \) is \( n \cdot x^{n-1} \). Applying this to our function, we take the derivative of \( -x^2 \). Here, \( n = 2 \) so, by the power rule, the derivative is \( -2x^{2-1} = -2x \).

This rule simplifies the differentiation process significantly, especially when dealing with polynomials. It can save a lot of time and prevent errors compared to finding derivatives by definition. It's one of the first and most important techniques you learn, as it underpins many differentiation problems.

The point-slope form of a linear equation is instrumental when writing the equation of a tangent line. Once we have the slope from differentiation, and a point through which the tangent line passes, we can use this form.

This form is expressed as \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the specific point on the line. In our exercise, the slope \( m \) was calculated as 2, and the point is \( (-1, 3) \). By plugging these into the point-slope form, we obtain:
\[ y - 3 = 2(x + 1) \]

• This equation forms the basis for the tangent line.
• It can be further simplified to the slope-intercept form, if needed.

In this instance, simplifying gives us \( y = 2x + 5 \). This equation is simple, showing both the slope of the tangent (2) and where it intersects the y-axis when extended in both directions (-5). Understanding how to apply the point-slope form helps in clearly defining lines tangentially related to curves at specific points.